\(\sqrt{16\sqrt{ }8\sqrt{ }4}\) bang bao nhieu
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`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`
`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`
`=root{3}{4(1-sqrt3)(1+sqrt3)}`
`=root{3}{4(1-3)}=-2`
`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`
`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`
`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`
`=root{3}{9}`
`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`
`=root{3}{(8sqrt5-16)(8sqrt5+16)}`
`=root{3}{320-256}`
`=root{3}{64}=4`
`b)root{3}{7-5sqrt2}-root{6}{8}`
`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`
`=root{3}{(1-sqrt2)^3}-sqrt2`
`=1-sqrt2-sqrt2=1-2sqrt2`
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:Áp dụng định lý cos ta có:
\(\cos A=\frac{AB^2+AC^2-BC^2}{2AB.AC}=\frac{-1}{2}\Rightarrow \widehat{A}=120^0\)
\(\cos B=\frac{BC^2+BA^2-AC^2}{2BC.BA}=\frac{-\sqrt{2}}{2}\Rightarrow \widehat{B}=45^0\)
\(\widehat{C}=180^0-(\widehat{A}+\widehat{B})=180^0-(120^0+45^0)=15^0\)
\(\widehat{ADB}=180^0-(\frac{\widehat{A}}{2}+\widehat{B})=180^0-(\frac{120^0}{2}+45^0)=75^0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{3}{\sqrt{n}+\sqrt{n+4}}=\frac{3}{4}.\frac{4}{\sqrt{n}+\sqrt{n+4}}=\frac{3}{4}.\frac{4\left(\sqrt{n+4}-\sqrt{n}\right)}{\left(\sqrt{n+4}+\sqrt{n}\right)\left(\sqrt{n+4}-\sqrt{n}\right)}\)
\(=\frac{3}{4}.\frac{4\left(\sqrt{n+4}-\sqrt{n}\right)}{n+4-n}=\frac{3}{4}.\frac{4\left(\sqrt{n+4}-\sqrt{n}\right)}{4}=\frac{3}{4}\left(\sqrt{n+4}-\sqrt{n}\right)\)
Áp dụng ta được :
\(\frac{3}{\sqrt{4}+\sqrt{8}}+\frac{3}{\sqrt{8}+\sqrt{12}}+\frac{3}{\sqrt{12}+\sqrt{16}}+...+\frac{3}{\sqrt{572}+\sqrt{576}}\)
\(=\frac{3}{4}\left(\sqrt{8}-\sqrt{4}+\sqrt{12}-\sqrt{8}+\sqrt{16}-\sqrt{12}+...+\sqrt{576}-\sqrt{572}\right)\)
\(=\frac{3}{4}\left(\sqrt{576}-\sqrt{4}\right)=\frac{3}{4}\left(24-4\right)=\frac{3}{4}.20=15\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
b: \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{6a^2}{24}}=\sqrt{\dfrac{a^2}{4}}=\dfrac{a}{2}\)
c: \(\sqrt{5a\cdot45a}-3a=-15a-3a=-18a\)